Stirling numbers of the first kind

s(4,2)=11

The image at right shows that ${\displaystyle \left[{4 \atop 2}\right]=11}$: the symmetric group on 4 objects has 3 permutations of the form

${\displaystyle (\bullet \bullet )(\bullet \bullet )}$ (having 2 orbits, each of size 2),

and 8 permutations of the form

${\displaystyle (\bullet \bullet \bullet )(\bullet )}$ (having 1 orbit of size 3 and 1 orbit of size 1).

The signs of the (signed) Stirling numbers of the first kind are predictable and depend on the parity of n − k. In particular,

${\displaystyle s(n,k)=(-1)^{n-k}\left[{n \atop k}\right].}$

Note that although

${\displaystyle \left[{0 \atop 0}\right]=1}$, we have ${\displaystyle \left[{n \atop 0}\right]=0}$ if n > 0

and

${\displaystyle \left[{0 \atop k}\right]=0}$ if k > 0, or more generally ${\displaystyle \left[{n \atop k}\right]=0}$ if k > n.

Also

${\displaystyle \left[{n \atop 1}\right]=(n-1)!,\quad \left[{n \atop n}\right]=1,\quad \left[{n \atop n-1}\right]={n \choose 2},}$

and

${\displaystyle \left[{n \atop n-2}\right]={\frac {1}{4}}(3n-1){n \choose 3}\quad {\mbox{ and }}\quad \left[{n \atop n-3}\right]={n \choose 2}{n \choose 4}.}$

Similar relationships involving the Stirling numbers hold for the Bernoulli polynomials. Many relations for the Stirling numbers shadow similar relations on the binomial coefficients. The study of these 'shadow relationships' is termed umbral calculus and culminates in the theory of Sheffer sequences. Generalizations of the Stirling numbers of both kinds to arbitrary complex-valued inputs may be extended through the relations of these triangles to the Stirling convolution polynomials.[1]

Since the Stirling numbers are the coefficients of a polynomial with roots 0, 1, ..., n − 1, one has by Vieta's formulas that

$${\displaystyle \left[{\begin{matrix}n\\n-k\end{matrix}}\right]=\sum _{0\leq i_{1}<i_{2}<\ldots <i_{k}<n}i_{1}i_{2}\cdots i_{k}.}$$

In other words, the Stirling numbers of the first kind are given by elementary symmetric polynomials evaluated at 0, 1, ..., n − 1.[2] In this form, the simple identities given above take the form

$${\displaystyle \left[{\begin{matrix}n\\n-1\end{matrix}}\right]=\sum _{i=0}^{n-1}i={\binom {n}{2}},}$$

$${\displaystyle \left[{\begin{matrix}n\\n-2\end{matrix}}\right]=\sum _{i=0}^{n-1}\sum _{j=0}^{i-1}ij={\frac {3n-1}{4}}{\binom {n}{3}},}$$

$${\displaystyle \left[{\begin{matrix}n\\n-3\end{matrix}}\right]=\sum _{i=0}^{n-1}\sum _{j=0}^{i-1}\sum _{k=0}^{j-1}ijk={\binom {n}{2}}{\binom {n}{4}},}$$

and so on.

One may produce alternative forms for the Stirling numbers of the first kind with a similar approach preceded by some algebraic manipulation: since

${\displaystyle (x+1)(x+2)\cdots (x+n-1)=(n-1)!\cdot (x+1)\left({\frac {x}{2}}+1\right)\cdots \left({\frac {x}{n-1}}+1\right),}$

it follows from Newton's formulas that one can expand the Stirling numbers of the first kind in terms of generalized harmonic numbers. This yields identities like

${\displaystyle \left[{n \atop 2}\right]=(n-1)!\;H_{n-1},}$

${\displaystyle \left[{n \atop 3}\right]={\frac {1}{2}}(n-1)!\left[(H_{n-1})^{2}-H_{n-1}^{(2)}\right]}$

${\displaystyle \left[{n \atop 4}\right]={\frac {1}{3!}}(n-1)!\left[(H_{n-1})^{3}-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right],}$

where H_n is the harmonic number ${\displaystyle H_{n}={\frac {1}{1}}+{\frac {1}{2}}+\ldots +{\frac {1}{n}}}$ and H_n^(m) is the generalized harmonic number

$${\displaystyle H_{n}^{(m)}={\frac {1}{1^{m}}}+{\frac {1}{2^{m}}}+\ldots +{\frac {1}{n^{m}}}.}$$

These relations can be generalized to give

${\displaystyle {\frac {1}{(n-1)!}}\left[{\begin{matrix}n\\k+1\end{matrix}}\right]=\sum _{i_{1}=1}^{n-1}\sum _{i_{2}=i_{1}+1}^{n-1}\cdots \sum _{i_{k}=i_{k-1}+1}^{n-1}{\frac {1}{i_{1}i_{2}\cdots i_{k}}}={\frac {w(n,k)}{k!}}}$

where w(n, m) is defined recursively in terms of the generalized harmonic numbers by

${\displaystyle w(n,m)=\delta _{m,0}+\sum _{k=0}^{m-1}(1-m)_{k}H_{n-1}^{(k+1)}w(n,m-1-k).}$

(Here δ is the Kronecker delta function and ${\displaystyle (m)_{k}}$ is the Pochhammer symbol.)[3]

For fixed ${\displaystyle n\geq 0}$ these weighted harmonic number expansions are generated by the generating function

${\displaystyle {\frac {1}{n!}}\left[{\begin{matrix}n+1\\k\end{matrix}}\right]=[x^{k}]\exp \left(\sum _{m\geq 1}{\frac {(-1)^{m-1}H_{n}^{(m)}}{m}}x^{m}\right),}$

where the notation ${\displaystyle [x^{k}]}$ means extraction of the coefficient of ${\displaystyle x^{k}}$ from the following formal power series (see the non-exponential Bell polynomials and section 3 of [4]).

More generally, sums related to these weighted harmonic number expansions of the Stirling numbers of the first kind can be defined through generalized zeta series transforms of generating functions.[5][6]

One can also "invert" the relations for these Stirling numbers given in terms of the ${\displaystyle k}$-order harmonic numbers to write the integer-order generalized harmonic numbers in terms of weighted sums of terms involving the Stirling numbers of the first kind. For example, when ${\displaystyle k=2,3}$ the second-order and third-order harmonic numbers are given by

${\displaystyle (n!)^{2}\cdot H_{n}^{(2)}=\left[{\begin{matrix}n+1\\2\end{matrix}}\right]^{2}-2\left[{\begin{matrix}n+1\\1\end{matrix}}\right]\left[{\begin{matrix}n+1\\3\end{matrix}}\right]}$

${\displaystyle (n!)^{3}\cdot H_{n}^{(3)}=\left[{\begin{matrix}n+1\\2\end{matrix}}\right]^{3}-3\left[{\begin{matrix}n+1\\1\end{matrix}}\right]\left[{\begin{matrix}n+1\\2\end{matrix}}\right]\left[{\begin{matrix}n+1\\3\end{matrix}}\right]+3\left[{\begin{matrix}n+1\\1\end{matrix}}\right]^{2}\left[{\begin{matrix}n+1\\4\end{matrix}}\right].}$

More generally, one can invert the Bell polynomial generating function for the Stirling numbers expanded in terms of the ${\displaystyle m}$-order harmonic numbers to obtain that for integers ${\displaystyle m\geq 2}$

${\displaystyle H_{n}^{(m)}=-m\times [x^{m}]\log \left(1+\sum _{k\geq 1}\left[{\begin{matrix}n+1\\k+1\end{matrix}}\right]{\frac {(-x)^{k}}{n!}}\right).}$

For all positive integer m and n, one has

$${\displaystyle (n+m)!=\sum _{k=0}^{n}\sum _{j=0}^{m}m^{\overline {n-k}}\left[{m \atop j}\right]{\binom {n}{k}}k!,}$$

where ${\displaystyle a^{\overline {b}}=a(a+1)\cdots (a+b-1)}$ is the rising factorial.[7] This formula is a dual of Spivey's result for the Bell numbers.[7]

Other related formulas involving the falling factorials, Stirling numbers of the first kind, and in some cases Stirling numbers of the second kind include the following:[8]

$${\displaystyle {\begin{aligned}n^{\underline {n-m}}&=\sum _{k}\left[{\begin{matrix}n+1\\k+1\end{matrix}}\right]\left\{{\begin{matrix}k\\m\end{matrix}}\right\}(-1)^{m-k}\\{\binom {n}{m}}(n-1)^{\underline {n-m}}&=\sum _{k}\left[{\begin{matrix}n\\k\end{matrix}}\right]\left\{{\begin{matrix}k\\m\end{matrix}}\right\}\\{\binom {n}{m}}&=\sum _{k}\left\{{\begin{matrix}n+1\\k+1\end{matrix}}\right\}\left[{\begin{matrix}k\\m\end{matrix}}\right](-1)^{m-k}\\\left[{\begin{matrix}n+1\\m+1\end{matrix}}\right]&=\sum _{0\leq k\leq n}\left[{\begin{matrix}k\\m\end{matrix}}\right]n^{\underline {n-k}}.\end{aligned}}}$$

For any pair of sequences, ${\displaystyle \{f_{n}\}}$ and ${\displaystyle \{g_{n}\}}$, related by a finite sum Stirling number formula given by

${\displaystyle g_{n}=\sum _{k=1}^{n}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}f_{k},}$

for all integers ${\displaystyle n\geq 0}$, we have a corresponding inversion formula for ${\displaystyle f_{n}}$ given by

${\displaystyle f_{n}=\sum _{k=1}^{n}\left[{\begin{matrix}n\\k\end{matrix}}\right](-1)^{n-k}g_{k}.}$

These inversion relations between the two sequences translate into functional equations between the sequence exponential generating functions given by the Stirling (generating function) transform as

${\displaystyle {\widehat {G}}(z)={\widehat {F}}\left(e^{z}-1\right)}$

and

${\displaystyle {\widehat {F}}(z)={\widehat {G}}\left(\log(1+z)\right).}$

The differential operators ${\displaystyle D=d/dx}$ and ${\displaystyle \vartheta =zD}$ are related by the following formulas for all integers ${\displaystyle n\geq 0}$:[9]

${\displaystyle \vartheta ^{n}=\sum _{k}S(n,k)z^{k}D^{k}}$

${\displaystyle z^{n}D^{n}=\sum _{k}s(n,k)\vartheta ^{k}}$

Another pair of "inversion" relations involving the Stirling numbers relate the forward differences and the ordinary ${\displaystyle n^{th}}$ derivatives of a function, ${\displaystyle f(x)}$, which is analytic for all ${\displaystyle x}$ by the formulas[10]

${\displaystyle {\frac {1}{k!}}{\frac {d^{k}}{dx^{k}}}f(x)=\sum _{n=k}^{\infty }{\frac {s(n,k)}{n!}}\Delta ^{n}f(x)}$

${\displaystyle {\frac {1}{k!}}\Delta ^{k}f(x)=\sum _{n=k}^{\infty }{\frac {S(n,k)}{n!}}{\frac {d^{n}}{dx^{n}}}f(x).}$

The following congruence identity may be proved via a generating function-based approach:[11]

${\displaystyle {\begin{aligned}\left[{\begin{matrix}n\\m\end{matrix}}\right]&\equiv {\binom {\lfloor n/2\rfloor }{m-\lceil n/2\rceil }}=[x^{m}]\left(x^{\lceil n/2\rceil }(x+1)^{\lfloor n/2\rfloor }\right)&&{\pmod {2}},\end{aligned}}}$

More recent results providing Jacobi-type J-fractions that generate the single factorial function and generalized factorial-related products lead to other new congruence results for the Stirling numbers of the first kind.[12] For example, working modulo ${\displaystyle 2}$ we can prove that

${\displaystyle {\begin{aligned}\left[{\begin{matrix}n\\1\end{matrix}}\right]&\equiv {\frac {2^{n}}{4}}[n\geq 2]+[n=1]&&{\pmod {2}}\\\left[{\begin{matrix}n\\2\end{matrix}}\right]&\equiv {\frac {3\cdot 2^{n}}{16}}(n-1)[n\geq 3]+[n=2]&&{\pmod {2}}\\\left[{\begin{matrix}n\\3\end{matrix}}\right]&\equiv 2^{n-7}(9n-20)(n-1)[n\geq 4]+[n=3]&&{\pmod {2}}\\\left[{\begin{matrix}n\\4\end{matrix}}\right]&\equiv 2^{n-9}(3n-10)(3n-7)(n-1)[n\geq 5]+[n=4]&&{\pmod {2}}\\\left[{\begin{matrix}n\\5\end{matrix}}\right]&\equiv 2^{n-13}(27n^{3}-279n^{2}+934n-1008)(n-1)[n\geq 6]+[n=5]&&{\pmod {2}}\\\left[{\begin{matrix}n\\6\end{matrix}}\right]&\equiv {\frac {2^{n-15}}{5}}(9n^{2}-71n+120)(3n-14)(3n-11)(n-1)[n\geq 7]+[n=6]&&{\pmod {2}},\end{aligned}}}$

and working modulo ${\displaystyle 3}$ we can similarly prove that

${\displaystyle {\begin{aligned}\left[{\begin{matrix}n\\1\end{matrix}}\right]&\equiv \sum \limits _{j=\pm 1}{\frac {1}{36}}\left(9-5j{\sqrt {3}}\right)\times \left(3+j{\sqrt {3}}\right)^{n}[n\geq 2]+[n=1]&&{\pmod {3}}\\\left[{\begin{matrix}n\\2\end{matrix}}\right]&\equiv \sum \limits _{j=\pm 1}{\frac {1}{216}}\left((44n-41)-(25n-24)\cdot j{\sqrt {3}}\right)\times \left(3+j{\sqrt {3}}\right)^{n}[n\geq 3]+[n=2]&&{\pmod {3}}\\\left[{\begin{matrix}n\\3\end{matrix}}\right]&\equiv \sum \limits _{j=\pm 1}{\frac {1}{15552}}\left((1299n^{2}-3837n+2412)-(745n^{2}-2217n+1418)\cdot j{\sqrt {3}}\right)\times \left(3+j{\sqrt {3}}\right)^{n}[n\geq 4]+[n=3]&&{\pmod {3}}\\\left[{\begin{matrix}n\\4\end{matrix}}\right]&\equiv \sum \limits _{j=\pm 1}{\frac {1}{179936}}{\bigl (}(6409n^{3}-383778n^{2}+70901n-37092)-(3690n^{3}-22374n^{2}+41088n-21708)\cdot j{\sqrt {3}}{\bigr )}\times \left(3+j{\sqrt {3}}\right)^{n}[n\geq 5]+[n=4]&&{\pmod {3}}.\end{aligned}}}$

More generally, for fixed integers ${\displaystyle h\geq 3}$ if we define the ordered roots

${\displaystyle \left(\omega _{h,i}\right)_{i=1}^{h-1}:=\left\{\omega _{j}:\sum _{i=0}^{h-1}{\binom {h-1}{i}}{\frac {h!}{(i+1)!}}(-\omega _{j})^{i}=0,\ 1\leq j<h\right\},}$

then we may expand congruences for these Stirling numbers defined as the coefficients

${\displaystyle \left[{\begin{matrix}n\\m\end{matrix}}\right]=[R^{m}]R(R+1)\cdots (R+n-1),}$

in the following form where the functions, ${\displaystyle p_{h,i}^{[m]}(n)}$, denote fixed polynomials of degree ${\displaystyle m}$ in ${\displaystyle n}$ for each ${\displaystyle h}$, ${\displaystyle m}$, and ${\displaystyle i}$:

${\displaystyle \left[{\begin{matrix}n\\m\end{matrix}}\right]=\left(\sum _{i=0}^{h-1}p_{h,i}^{[m]}(n)\times \omega _{h,i}^{n}\right)[n>m]+[n=m]\qquad {\pmod {h}},}$

Section 6.2 of the reference cited above provides more explicit expansions related to these congruences for the ${\displaystyle r}$-order harmonic numbers and for the generalized factorial products, ${\displaystyle p_{n}(\alpha ,R):=R(R+\alpha )\cdots (R+(n-1)\alpha )}$. In the previous examples, the notation ${\displaystyle [{\mathtt {cond}}]}$ denotes Iverson's convention.

A variety of identities may be derived by manipulating the generating function:

${\displaystyle H(z,u)=(1+z)^{u}=\sum _{n=0}^{\infty }{u \choose n}z^{n}=\sum _{n=0}^{\infty }{\frac {z^{n}}{n!}}\sum _{k=0}^{n}s(n,k)u^{k}=\sum _{k=0}^{\infty }u^{k}\sum _{n=k}^{\infty }{\frac {z^{n}}{n!}}s(n,k).}$

Using the equality

${\displaystyle (1+z)^{u}=e^{u\log(1+z)}=\sum _{k=0}^{\infty }(\log(1+z))^{k}{\frac {u^{k}}{k!}},}$

it follows that

${\displaystyle \sum _{n=k}^{\infty }(-1)^{n-k}\left[{n \atop k}\right]{\frac {z^{n}}{n!}}={\frac {\left(\log(1+z)\right)^{k}}{k!}}.}$

(This identity is valid for formal power series, and the sum converges in the complex plane for |z| < 1.) Other identities arise by exchanging the order of summation, taking derivatives, making substitutions for z or u, etc. For example, we may derive:[13]

${\displaystyle (1-z)^{-u}=\sum _{k=0}^{\infty }u^{k}\sum _{n=k}^{\infty }{\frac {z^{n}}{n!}}\left[{n \atop k}\right]=e^{u\log(1/(1-z))}}$

and

${\displaystyle {\frac {\log ^{m}(1+z)}{1+z}}=m!\sum _{k=0}^{\infty }{\frac {s(k+1,m+1)\,z^{k}}{k!}},\qquad m=1,2,3,\ldots \quad |z|<1}$

or

${\displaystyle \sum _{n=i}^{\infty }{\frac {\left[{n \atop i}\right]}{n\,(n!)}}=\zeta (i+1),\qquad i=1,2,3,\ldots }$

and

${\displaystyle \sum _{n=i}^{\infty }{\frac {\left[{n \atop i}\right]}{n\,(v)_{n}}}=\zeta (i+1,v),\qquad i=1,2,3,\ldots \quad \Re (v)>0}$

where ${\displaystyle \zeta (k)}$ and ${\displaystyle \zeta (k,v)}$ are the Riemann zeta function and the Hurwitz zeta function respectively, and even evaluate this integral

${\displaystyle \int _{0}^{1}{\frac {\log ^{z}(1-x)}{x^{k}}}\,dx={\frac {(-1)^{z}\Gamma (z+1)}{(k-1)!}}\sum _{r=1}^{k-1}s(k-1,r)\sum _{m=0}^{r}{\binom {r}{m}}(k-2)^{r-m}\zeta (z+1-m),\qquad \Re (z)>k-1,\quad k=3,4,5,\ldots }$

where ${\displaystyle \Gamma (z)}$ is the Gamma function. There also exist more complicated expressions for the zeta-functions involving the Stirling numbers. One, for example, has

${\displaystyle {\frac {k!}{(s-k)_{k}}}\sum _{n=0}^{\infty }{\frac {1}{(n+k)!}}\left[{n+k \atop n}\right]\sum _{l=0}^{n+k-1}\!(-1)^{l}{\binom {n+k-1}{l}}(l+v)^{k-s}=\zeta (s,v),\quad k=1,2,3,\ldots }$

This series generalizes Hasse's series for the Hurwitz zeta-function (we obtain Hasse's series by setting k=1).[14][15]

The next estimate given in terms of the Euler gamma constant applies:[16]

${\displaystyle \left[{\begin{matrix}n+1\\k+1\end{matrix}}\right]\sim {\frac {n!}{k!}}\left(\gamma +\ln n\right)^{k},\ {\text{ uniformly for }}k=o(\ln n).}$

For fixed ${\displaystyle n}$ we have the following estimate as ${\displaystyle k\longrightarrow \infty }$:

${\displaystyle \left[{\begin{matrix}n+k\\k\end{matrix}}\right]\sim {\frac {k^{2n}}{2^{n}n!}}.}$

We can also apply the saddle point asymptotic methods from Temme's article [17] to obtain other estimates for the Stirling numbers of the first kind. These estimates are more involved and complicated to state. Nonetheless, we provide an example. In particular, we define the log gamma function, ${\displaystyle \phi (x)}$, whose higher-order derivatives are given in terms of polygamma functions as

${\displaystyle {\begin{aligned}\phi (x)&=\ln \left[(1+1)(x+2)\cdots (x+n)\right]-m\ln x\\&=\ln \Gamma (x+n+1)-\ln \Gamma (x+1)-m\ln x\\\phi ^{\prime }(x)&=\psi (x+n+1)-\psi (x+1)-m/x,\end{aligned}}}$

where we consider the (unique) saddle point ${\displaystyle x_{0}}$ of the function to be the solution of ${\displaystyle \phi ^{\prime }(x)=0}$ when ${\displaystyle 1\leq m\leq n}$. Then for ${\displaystyle t_{0}=m/(n-m)}$ and the constants

${\displaystyle B=\phi (x_{0})-n\ln(1+t_{0})+m\ln t_{0}}$

${\displaystyle g(t_{0})={\frac {1}{x_{0}}}{\sqrt {\frac {m(n-m)}{n\phi ^{\prime \prime }(x_{0})}}},}$

we have the following asymptotic estimate as ${\displaystyle n\longrightarrow \infty }$:

${\displaystyle \left[{\begin{matrix}n+1\\m+1\end{matrix}}\right]\sim e^{B}g(t_{0}){\binom {n}{m}}.}$

Since permutations are partitioned by number of cycles, one has

${\displaystyle \sum _{k=0}^{n}\left[{n \atop k}\right]=n!}$

The identity

${\displaystyle \sum _{p=k}^{n}{\left[{n \atop p}\right]{\binom {p}{k}}}=\left[{n+1 \atop k+1}\right]}$

can be proved by the techniques on the page Stirling numbers and exponential generating functions.

The table in section 6.1 of Concrete Mathematics provides a plethora of generalized forms of finite sums involving the Stirling numbers. Several particular finite sums relevant to this article include

${\displaystyle {\begin{aligned}\left[{\begin{matrix}n\\m\end{matrix}}\right]&=\sum _{k}\left[{\begin{matrix}n+1\\k+1\end{matrix}}\right]{\binom {k}{m}}(-1)^{m-k}\\\left[{\begin{matrix}n+1\\m+1\end{matrix}}\right]&=\sum _{k=0}^{n}\left[{\begin{matrix}k\\m\end{matrix}}\right]{\frac {n!}{k!}}\\\left[{\begin{matrix}m+n+1\\m\end{matrix}}\right]&=\sum _{k=0}^{m}(n+k)\left[{\begin{matrix}n+k\\k\end{matrix}}\right]\\\left[{\begin{matrix}n\\l+m\end{matrix}}\right]{\binom {l+m}{l}}&=\sum _{k}\left[{\begin{matrix}k\\l\end{matrix}}\right]\left[{\begin{matrix}n-k\\m\end{matrix}}\right]{\binom {n}{k}}.\end{aligned}}}$

Other finite sum identities involving the signed Stirling numbers of the first kind found, for example, in the NIST Handbook of Mathematical Functions (Section 26.8) include the following sums:[18]

${\displaystyle {\begin{aligned}{\binom {k}{h}}s(n,k)&=\sum _{j=k-h}^{n-h}{\binom {n}{j}}s(n-j,h)s(j,k-h)\\s(n+1,k+1)&=n!\times \sum _{j=k}^{n}{\frac {(-1)^{n-j}}{j!}}s(j,k)\\s(n,n-k)&=\sum _{j=0}^{k}(-1)^{j}{\binom {n-1+j}{k+j}}{\binom {n+k}{k-j}}S(k+j,j)\\s(n,k)&=\sum _{j=k}^{n}s(n+1,j+1)n^{j-k}.\end{aligned}}}$

Additionally, if we define the second-order Eulerian numbers by the triangular recurrence relation [19]

${\displaystyle E_{2}(n,k)=(k+1)E_{2}(n-1,k)+(2n-1-k)E_{2}(n-1,k-1)+\delta _{n,0}\delta _{k,0},}$

we arrive at the following identity related to the form of the Stirling convolution polynomials which can be employed to generalize both Stirling number triangles to arbitrary real, or complex-valued, values of the input ${\displaystyle x}$:

${\displaystyle \left[{\begin{matrix}x\\x-n\end{matrix}}\right]=\sum _{k}E_{2}(n,k){\binom {x+k}{2n}}.}$

Particular expansions of the previous identity lead to the following identities expanding the Stirling numbers of the first kind for the first few small values of ${\displaystyle n:=1,2,3}$:

${\displaystyle {\begin{aligned}\left[{\begin{matrix}x\\x-1\end{matrix}}\right]&={\binom {x}{2}}\\\left[{\begin{matrix}x\\x-2\end{matrix}}\right]&={\binom {x}{4}}+2{\binom {x+1}{4}}\\\left[{\begin{matrix}x\\x-3\end{matrix}}\right]&={\binom {x}{6}}+8{\binom {x+1}{6}}+6{\binom {x+2}{6}}.\end{aligned}}}$

Software tools for working with finite sums involving Stirling numbers and Eulerian numbers are provided by the RISC Stirling.m package utilities in Mathematica. Other software packages for guessing formulas for sequences (and polynomial sequence sums) involving Stirling numbers and other special triangles is available for both Mathematica and Sage here and here, respectively.[20]

Furthermore, infinite series involving the finite sums with the Stirling numbers often lead to the special functions. For example[13][21]

${\displaystyle \sum _{n=1}^{\infty }\!{\frac {(-1)^{n-1}}{n}}\cdot {\frac {1}{n!}}\sum _{l=1}^{n}\!{\frac {s(n,l)}{l+k}}=\sum _{l=2}^{\infty }\!{\frac {(-1)^{l}\cdot \zeta (l)}{l+k-1}}={\frac {1}{k-1}}-{\frac {\ln 2\pi }{k}}+{\frac {\gamma }{2}}+\!\!\!\!\sum _{l=1}^{\lfloor {\frac {1}{2}}(k-1)\rfloor }\!\!\!\!(-1)^{l}{\binom {k-1}{2l-1}}{\frac {(2l)!\cdot \zeta '(2l)}{l\cdot (2\pi )^{2l}}}+\!\!\sum _{l=1}^{\lfloor {\frac {1}{2}}k\rfloor -1}\!\!(-1)^{l}{\binom {k-1}{2l}}{\frac {(2l)!\cdot \zeta (2l+1)}{2\cdot (2\pi )^{2l}}}}$

or

${\displaystyle \ln \Gamma (z)=\left(z-{\frac {1}{2}}\right)\!\ln z-z+{\frac {1}{2}}\ln 2\pi +{\frac {1}{\pi }}\sum _{n=1}^{\infty }{\frac {1}{n\cdot n!}}\!\sum _{l=0}^{\lfloor {\frac {1}{2}}n\rfloor }\!{\frac {(-1)^{l}(2l)!}{(2\pi z)^{2l+1}}}\left[{n \atop 2l+1}\right]}$

and

${\displaystyle \Psi (z)=\ln z-{\frac {1}{2z}}-{\frac {1}{\pi z}}\sum _{n=1}^{\infty }{\frac {1}{n\cdot n!}}\!\sum _{l=0}^{\lfloor {\frac {1}{2}}n\rfloor }\!{\frac {(-1)^{l}(2l+1)!}{(2\pi z)^{2l+1}}}\left[{n \atop 2l+1}\right]}$

or even

${\displaystyle \gamma _{m}={\frac {1}{2}}\delta _{m,0}+{\frac {(-1)^{m}m!}{\pi }}\!\sum _{n=1}^{\infty }{\frac {1}{n\cdot n!}}\!\sum _{k=0}^{\lfloor {\frac {1}{2}}n\rfloor }{\frac {(-1)^{k}}{(2\pi )^{2k+1}}}\left[{2k+2 \atop m+1}\right]\left[{n \atop 2k+1}\right]\,}$

where γ_m are the Stieltjes constants and δ_m,0 represents the Kronecker delta function.

The Stirling number s(n,n-p) can be found from the formula[22]

${\displaystyle {\begin{aligned}s(n,n-p)&={\frac {1}{(n-p-1)!}}\sum _{0\leq k_{1},\ldots ,k_{p}:\sum _{1}^{p}mk_{m}=p}(-1)^{K}{\frac {(n+K-1)!}{k_{1}!k_{2}!\cdots k_{p}!~2!^{k_{1}}3!^{k_{2}}\cdots (p+1)!^{k_{p}}}},\end{aligned}}}$

where ${\displaystyle K=k_{1}+\cdots +k_{p}.}$ The sum is a sum over all partitions of p.

Another exact nested sum expansion for these Stirling numbers is computed by elementary symmetric polynomials corresponding to the coefficients in ${\displaystyle x}$ of a product of the form ${\displaystyle (1+c_{1}x)\cdots (1+c_{n-1}x)}$. In particular, we see that

${\displaystyle {\begin{aligned}\left[{\begin{matrix}n\\k+1\end{matrix}}\right]&=[x^{k}](x+1)(x+2)\cdots (x+n-1)=(n-1)!\cdot [x^{k}](x+1)\left({\frac {x}{2}}+1\right)\cdots \left({\frac {x}{n-1}}+1\right)\\&=\sum _{1\leq i_{1}<\cdots <i_{k}<n}{\frac {(n-1)!}{i_{1}\cdots i_{k}}}.\end{aligned}}}$

Newton's identities combined with the above expansions may be used to give an alternate proof of the weighted expansions involving the generalized harmonic numbers already noted above.

Another explicit formula for reciprocal powers of n is given by the following identity for integers ${\displaystyle p\geq 2}$:[23]

${\displaystyle {\frac {1}{n^{p}}}={\frac {1}{n^{p-1}}}+{\frac {(-1)^{p-1}}{n!}}\left({\begin{bmatrix}n\\p\end{bmatrix}}+{\begin{bmatrix}n\\p-1\end{bmatrix}}\right)+{\frac {(-1)^{p}}{n\cdot n!}}{\begin{bmatrix}n+1\\p\end{bmatrix}}+\sum _{j=0}^{p-3}{\begin{bmatrix}n+1\\j+2\end{bmatrix}}{\frac {(-1)^{j+1}(n-1)}{n^{p-1-j}\cdot n!}}.}$